CHAPTER V

TEMPERATURE AND HEAT

A. TEMPERATUREThe tool used to measure the temperature a thermometer called there are three kind of thermometer namely : Celsius, reamer, Fahrenheit.

100 Scale80 Scale 180 Scale

1000

320

800

R

2120

00 00

CMelting point

F

C = thermometer Celsius (boiling point = 1000 C )

R = thermometer Reamer (boiling point = 800 R )

F= thermometer Fahrenheit (boiling point = 2120 F )

K= thermometer Kelvin (boiling point= 373 K)

Boiling point

K

273

373

100 Scale

The Formula :tC : tR : ( tF – 32 ): (tK – 273) = 100 : 80 : 180: 100

tC : tR : ( tF – 32 ) : (tK - 273) = 5 : 4 : 9 : 5

The unit of absolute temperature is Kelvin.

T = tC + 273

The general formula:

mYbY

mY

mXbX

mX

ttttY

ttttX

tX = temperatur of thermometer X

tmX= melting point of thermometer X

tbX = boiling point of thermometer X

tY = temperatur of thermometer YtmY= melting point of thermometer YtbY = boiling point of thermometer Y

Problems.1. Scale of the Celsius thermometer show 500.

What is the scale of the thermometer :a. Kelvinb. Reamurc. Fahrenheit

2. a. 73 K = ….oF b. 40 oR= …OK c. 45 OF=…oK3. If X scale based on Celsius scale and have melting point 100 and

boiling point 1400. determine Celcius scale if X scale 400.4. Thermometer B have a melting point – 50 and boiling point 1150.

determine the temperature of B thermometer, if the Fahrenheit scale is 1130 F.

5. Find the Celsius scale which is equal to the Fahrenheit scale. Find also the reamur scale which is equal to the Fahrenheit scale.

6. Find the Fahrenheit scale which is bigger 200 than the reamer scale. Find also the Celsius scale which is smaller 100 than the Fahrenheit.

B. HEATIs the energy transferred from one object to the other of the temperature differences.The amount of energy absorbed or released of an object can be state with the formula:

Q = m . c . t

C = m . c

Q = C . t

Q = heat (J)

m = mass (kg)

c = specific heat of substance (J/kg K)

C = heat capacity (J/kg)

t = temperature changes (K)1 cal = 4,2 J

1 cal/g.oC = 4200 J/kg K

Problems :1. An amount of 3.106 j heat is given to the 4 kg aluminum box with the

temperature of 300 C ( c = 900 J / kg 0c ). Find the end temperature of the box.2. An object of 400 g with a specific heat of 3600 J / kg 0c is heated so the

temperature raise from 250C to 1000C.Find : a. Heat capacity. b. the absorb heat.

3. An aluminum foil of 200 g with a temperature of 900C is put inside 100 g water of 200C. Suppose that there is no missing heat. cal = 900 J / kg 0C, cw = 4200 J / kg 0C find the end temperature of the mixture.

4. A 0,5 kg tin of 1000C plunge into 0,2 kg water of 200C in an aluminum calorie meter of 0,1 kg. The end of the temperature is 240C. cAl = 900 J / kg k, ca = 4200 J / kg k.

find the tin specific heat ? 5. Ice of 50 g at -10oC is put on a 100 g hot plate of glass at 300oC, find the final

temperature of the system. (cice= 0,5 cal/g oC, Lice= 80 cal/g, cwater= 1 cal/g oC, and cglass= 0,2 cal/g oC)

C. PHASE CHANGES OF SUBSTANCES Graphic :

Gas

Solid (padat)

Liquid (Cair)

Q = m . L L = Latent Heat (J/kg)

Q1

00

Q2 Q3

1000C

Q4

1000C

00

Solid absorb heat energy of Q1 = m . cs.Δt

Melting absorb heat energy of Q2 = m . Lm

Liquid absorb heat energy of Q3 = m . cl . Δt

Boiling absorb heat energy of Q4 = m . Lb

gas absorb heat energy of Q5 = m. cg . Δt

notice: Lm = Melting latent heat

Lb = Boiling latent heat

The changes of states especially for water from solid ( ice ) to gas ( steam ) caused by giving heat energy can explain as in the graph below :

The problem of heat is on the black’s principle which is state that the amount of the absorb heat is the same with the release heat.

Heat lost by hot object = Heat gained by cold object

In symbols,

C. BLACK’S PRINCIPLE

Q release = Q absorb

Problems1. Find the heat energy to change 4 Kg of ice – 100C to gas of 1000C

ci = 2100 J/Kg k Le = 2256.103 J / kg

cw= 4200 J / kg k Lf = 334.103 J / kg

2. 25 gr of solid matter absorb heat energy of 200 J/s so it change all to gas in the graph :

020 60 75 90

A

time ( minute )

Find specific latent heat of fusion and evaporation!

B C

D E

F

3. 400 g ice of -40C put into 800 g water of 600C so the ice all melted. Find the final temperature of the mixture.

4. A glass tube (m=300g) of specific heat 0,2 cal / g0C filled with ice of 25 g -20C. Then water of 150 g 400C put into the tube so that all the ice are melted.Find the final temperature of the mixture.

D. Substance ExpansionAny substance which increase the temperature will be expanded, exceptions on water from 0°C to 4°C which will we reduce in volumes. This stronger feature is called water anomaly.

I. Solid Expansion Linear ExpansionAny solids with elongation ( eg.wire ) when it heated the length will be in creased.The elongations of the expansion can be found wit the formulate :

= Length elongation

= linier expansion coefficient

= Temperature

= Initial length

= Final length

tt

t) (1 ott .

o

Problems1. Length of a railway is made of steel is 20 m in 100C. Find the

length of the railway when the temperature is 400C.(α = 11.10-6/C)2. Two railway, each of then 15 m length will be set. If the minimum

and maximum temperature on that place is 250 C and 400 C, find the minimum distance two railway to be set. (α = 11.10-

6/C 3. A bar of steel (l = 2m) increase the length 2 mm when heated

arrive the temperature increase 600C. Find the length expansion of the steel when the temperature increase 200C.

II. Area ExpansionA plate when heated will be increase in area. The area expansion can be found by the formula :

= Area expansion= Area expansion coeff= Initial area= final area

21

tt

t

t

Problems1. A plaited of aluminum with its length of 20 cm and wide

12 cm heated from 200C to 500C.Find the final area of the plate.(α= 24.10-6/C)

2. A wheel of steel of 200C have an inside radius of 200 mm will be set to a wooden wheel of radius 202 mm it the linear expansion coefficient the steel 1,25.10-5/0C.Find the final temperature so that the wheel of steel will be set up to the wooden wheel precisely.

III. Liquids ExpansionLiquids when heated will increase in

volum,exceptionnally of water. When water heated from 00C to 40C the volume will reduce.

The liquids expansion can be found from the formula :

= Volume expansion= Volume expansion coefficient= Final volume= Initial volume

3

1VV

VV

γγ

γtt

t

V

γtVoV

Problem:1. The bar of steel of 20 cm x 5 cm x 3 cm temperature

400C. The bar is heated so it’s final volume 300,54 cm3. If the linear expansion coefficient of steel 15.10-6/0C. Find the final temperature the bar heated !

2. A bottle have a volume of 400 cm3 filled with water of 400C. Then the bottle heated to 600C. Find the water spilled out if the bottle expansion: (α glass= 9.10-6/oC, γ water= 2,1.10-4/oC)a. Neglectedb. Counted,

IV. Gas ExpansionWhen a gas in a closed room the temperature remain constant, according to Boyle, the changes of pressure and volumes is always constant.

According to Gay-Lussac, if the pressure is constant we have the relationship as follow :

constan P.V

konstan TV

If the formula (1) and (2) combine, the formula is know as Boyle-Gay Lussac formula.

= First Pressure ( atm,N/m2 )

= Last Pressure ( atm,N/m2 )= First Volume ( cm3,m3 )= Last Volume ( cm3,m3 )= First Temperature = Last Temperature 1atm = 76 cm Hg

konstan T

PV

2

22

1

11

TVP

TVP

1P

1V2V1T2T

2P

Problems1. A gas of H2 in a tube of 270C and volume 6 L, have the

pressure of 2 atm. Find the volume if the gas heated to 1770C in pressure constant.

2. A tube with it’s volume 8 L filled with gas of O22. The gas is heated to constant temperature so the pressure change to 3 times as before. Find the percentage of the volume changes ?.

E. Heat Transfer.Heat can travel in three ways :1. Conduction

Conduction is the flow of heat through matter without the changes of the particles of the medium.eg : heat transfer on a bar of iron.The amount of the heat energy conduction per second can be found in the formula

Q = Heat energyt = TimeK = thermal conductivityA = Aread = Length of the rod T = Temperature Changes

Two rods of different type join together will follow the rules that the heat energy transfer per second on the two rods have the same amount as in the formula below :

dTAK

tQ

..

Q1 Q2 =

t t

2. ConvectionConvection is the flow at heat with the changes of the particles of the medium. eg : heat transfer of water heating.The amount of the heat energy per second is :

3. RadiationRadiation is the flow of heat in the form of electromagnetic waves ( no medium melded ). The amount of energy radioactive is determined kg Stefan Bolliztman :

Q = Radiation energy A = Area of surface t = time T = Temperature σ = 5,67.10-8 watt / m2k4 e = Emisivity

TAhtQ

..h = Convection coefficient

4 etQ

Black body is an object that are able to amity / absorb of heat energy perfectly.If the temperature of the body is different we than the surroundings, if follows :

Problems1. An air conditional room have a glass window of 4 m2 area end the width of 2 mm. if

the temperature of the surface inside is 200C end outside 300C, find the heat energy per second.( kglass= 0,8 J / m.s.oC )

Two rods of P and Q jointed. If Kp = ½ KQ and AC = 2 CB, find the temperature of C.

3. An object of 2 m2 300C put in on a room of 200C find the heat energy emit from the object through convection in 5 minutes.

).( 42

41 e

tQ

P Q

A

400C

C B

1100C

2.

h = Convection coefficient =8 J/s.m2.oC

4. A string of wire on a filament lamp has an area of 100 mm2 11270C. It the wire supposed to be a black body find :a. The heat energy radiate each second.b. the electric current flow if the lamp connected to 220 V.

5. An object has an area 10 cm2 radiates on 5270C in a room of 1270C. The object emisivity of 0,8.Find then energy radiates in 1 minute heat.